# Circles

• Apr 11th 2014, 08:50 PM
xSummer
Circles
3. A small radio transmitter broadcasts in a 50 mile radius. If you drive along a straight line from a city 60 miles north of the transmitter to a second city 70 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

I am completely lost on how I'm supposed to go about solving this problem. The solution was provided with brief steps but I still couldn't wrap my head around it. Thank you!

-Summer

• Apr 11th 2014, 10:48 PM
romsek
Re: Circles
Quote:

Originally Posted by xSummer
3. A small radio transmitter broadcasts in a 50 mile radius. If you drive along a straight line from a city 60 miles north of the transmitter to a second city 70 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

I am completely lost on how I'm supposed to go about solving this problem. The solution was provided with brief steps but I still couldn't wrap my head around it. Thank you!

-Summer

Attachment 30672

You can see from the image that you intersect the circle of radio range in two spots.

Let the tower be at (0,0). City 1 is then at (0,60) and city 2 at (70,0).

Consider the line connecting the two cities.

The slope of this line is $m=\dfrac{(0-70)}{(60-0)}=- \dfrac 7 6$

The equation of the line is thus

$(y-60) = -\dfrac 7 6 x \Rightarrow y = -\dfrac 7 6 x$

In the quadrant where the trip takes place the equation for the circle of radio range is

$y = \sqrt{50^2 - x^2}=\sqrt{2500 - x^2}$

Now just solve for the two intersections

$\sqrt{2500 - x^2}=-\dfrac 7 6 x$

can you do this? You'l get two solutions corresponding to the two point on the circle. Find the distance between them. Find the distance between the cities. Take the ratio of these two distances and you're done.

The distance between two points is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
• Apr 12th 2014, 10:27 AM
romsek
Re: Circles
Quote:

Originally Posted by romsek
Attachment 30672

You can see from the image that you intersect the circle of radio range in two spots.

Let the tower be at (0,0). City 1 is then at (0,60) and city 2 at (70,0).

Consider the line connecting the two cities.

The slope of this line is $m=\dfrac{(0-70)}{(60-0)}=- \dfrac 7 6$

The equation of the line is thus

$(y-60) = -\dfrac 7 6 x \Rightarrow y = -\dfrac 7 6 x$

Typo here. What's meant is

$y=-\dfrac 7 6 x + 60$

Quote:

In the quadrant where the trip takes place the equation for the circle of radio range is

$y = \sqrt{50^2 - x^2}=\sqrt{2500 - x^2}$

Now just solve for the two intersections

$\sqrt{2500 - x^2}=-\dfrac 7 6 x$
and again what's meant is

$\sqrt{2500 - x^2}=-\dfrac 7 6 x + 60$

Quote:

can you do this? You'l get two solutions corresponding to the two point on the circle. Find the distance between them. Find the distance between the cities. Take the ratio of these two distances and you're done.

The distance between two points is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
• Apr 12th 2014, 11:52 AM
Soroban
Re: Circles
Hello, xSummer!

Did you make a sketch?

Quote:

If you drive along a straight line from a city 60 miles north of the transmitter
to a second city 70 miles east of the transmitter, during how much of the drive
will you pick up a signal from the transmitter?

On your graph, draw a circle centered at the Origin with radius 50.
Draw a line from (0,60) to (70,0).
Note where the line intersects the circle, points A and B.

• Apr 12th 2014, 04:20 PM
xSummer
Re: Circles
Hello Soroban,

Oops. I labeled points A & B incorrectly!

-Summer
• Apr 12th 2014, 04:24 PM
xSummer
Re: Circles
Quote:

Originally Posted by romsek
Typo here. What's meant is

$y=-\dfrac 7 6 x + 60$

Apologies but I'm only following you up till this point? Where did you get this second portion?

and again what's meant is

$\sqrt{2500 - x^2}=-\dfrac 7 6 x + 60$

-Summer
• Apr 12th 2014, 04:38 PM
romsek
Re: Circles
Quote:

Originally Posted by xSummer
-Summer

the equation for the 50 mile radius circle is given by $x^2 + y^2 = 50^2$

solving this for y in the first quadrant we get $y=\sqrt{2500 - x^2}$

Now set this equal to the formula for the line joining the two cities.

$\sqrt{2500 - x^2}=-\dfrac 7 6 x + 60$
• Apr 12th 2014, 05:17 PM
xSummer
Re: Circles
Romsek,

Oh, I see! Though I'm getting two really awkward numbers... Help?

(Also.. since I've been spending quite a bit of time with this problem I think I might've figured out another way to go about the problem:

Since the eq of the circle is: x^2 + y^2 = 50^2
And the equation of the line is -6/7x+60 which I would plug into the above equation?
(I believe you put -7/6 as the slope, it wouldn't be -6/7)

Though what I end up getting when I plug in is giving me a total different equation (far different then what the textbook calculates which is: x^2+(60-(60/70x))^2=50^2
• Apr 12th 2014, 05:44 PM
romsek
Re: Circles
Quote:

Originally Posted by xSummer
Romsek,

Oh, I see! Though I'm getting two really awkward numbers... Help?

(Also.. since I've been spending quite a bit of time with this problem I think I might've figured out another way to go about the problem:

Since the eq of the circle is: x^2 + y^2 = 50^2
And the equation of the line is -6/7x+60 which I would plug into the above equation?
(I believe you put -7/6 as the slope, it wouldn't be -6/7)

Though what I end up getting when I plug in is giving me a total different equation (far different then what the textbook calculates which is: x^2+(60-(60/70x))^2=50^2

oh bloody hell you're right. It should be -6/7. The answers aren't that different.

I get (14,48) and (45.2941, 21,1765) for the 2 points of intersection. What do you get?
• Apr 19th 2014, 06:23 PM
xSummer
Re: Circles
I'm so sorry for my late response! :( A lot came up. I figured it out! Thank you so much for your help.